Three circles each of radius 3.5 cm are drawn in such a way that each of them touches the other two. Find the area enclosed between these circles.

#### Solution

The three circles are drawn in such a way that each of them touches the other two.

So, by joining the centers of the three circles, we get,

AB = BC = CA = 2(Radius) = 7 cm

Therefore, triangle ABC is an equilateral triangle with each side 7 cm.

∴ Area of the triangle = `(sqrt(3)/4) xx a^2`

Where a is the side of the triangle.

= `(sqrt(3)/4) xx (7)^2`

= `49/9 sqrt(3) cm^2`

= 21.2176 cm^{2}

Now, Central angle of each sector = = 60° `((60π)/180)`

= `pi/3` radians

Thus, area of each sector = `(1/2)` r^{2}θ

= `(1/2) xx (3.5)^2 xx (pi/3)`

= `12.25 xx 22/(7 xx 6)`

= 6.4167 cm^{2}

Total area of three sectors = 3 × 6.4167 = 19.25 cm^{2}

∴ Area enclosed between three circles = Area of triangle ABC – Area of the three sectors

= 21.2176 – 19.25

= 1.9676 cm^{2}

Hence, the required area enclosed between these circles is 1.967 cm^{2} (approx.).